Question

Two 75 W (120 V) lightbulbs are wired in series, then the combination is connected to a 120 V supply. Part A How much power is dissipated by each bulb

Answer 1

Answer:

300 W

Explanation:

power of each bulb P = 75 W

voltage in the circuit = 120 V

we know that electrical power P = IV    ....1

and V = IR

we can also say that I = V/R

substituting for I in  equation 1, we have

P = $V^{2}/R$    ....2

The total total power in the circuit = 75 x 2 = 150 W

from equation 2, we have

150 = $120^{2} /R$

R = $120^{2}/150$ = 96 Ω    this is the resistance of the whole circuit.

This resistance is due to the two light bulbs, for each light bulb since they are arranged in series

R = 96/2 = 48 Ω

From P =  $V^{2}/R$  

for each light bulb, power is

P = $120^{2} /48$ = 300 W